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I’m currently getting ready for an Electronic Systems Engineering exam, and going through past paper questions.

I need help with a few questions regarding the following current limited regulator. enter image description here The questions:

  1. Find the zener voltage of Dz if the required regulated output is 12V.
  2. Calculate the value of R for safe operation if the input voltage is 15V and the power rating of Dz is 0.5W.

My answers: enter image description here

My questions to you:

  1. Have I approached the first question correctly? Can you please confirm my answer?
  2. In the second question: what is a “safe” current for the zener? Of course it should be less than the maximum current found from its power rating. But how much less? (I just randomly assumed 30mA)
K. Jayamanne
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    On the first question, I would assume no load current, so no drop across Rd. On the second one, you know the zener voltage and the max power, P=IV gives you the max current. Then take 15V-Vz to get the voltage across the resistor...you know the voltage and current, R=E/I. – Cristobol Polychronopolis Jan 04 '19 at 16:44
  • I would add to Cristobal's comment; assume negligible base current in Q1 when calculating the resistor. – Peter Smith Jan 04 '19 at 16:46
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    For an exam question I would suggest using the minimum resistor value from the stated zener dissipation, and adding a note about a real design using a higher value. Your answer is better in reality, of course, especially if you consider the effect of tolerances of input voltage and zener voltage. With a 2V drop, a 5% error in each could lead to a 62% increase in Zener dissipation, so a 300mW nominal dissipation would lead to ~500mW worst-case (not counting resistor tolerance). But don't waste time on this kind of thing during an exam. – Spehro Pefhany Jan 04 '19 at 18:42
  • @SpehroPefhany the minimum R for 5W into Zener means the current limiter burns up the R , who cares about protecting the Zener. Which uses less than the R in current limit mode!! Not prudent. With a current gain of 100 using a ratio of 50 means the R loss is only 2 % more in current limit mode or using a ratio of 30 3.3% wasted power. – Tony Stewart EE75 Jan 05 '19 at 06:26
  • From 1.6 V/30 mA, how do you get 80 Ω? – greybeard Mar 21 '23 at 07:57

1 Answers1

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The bigger question is what V tolerance and max power is used .

  • the Power rating of R needed for a short circuit output.

  • This will limit your power output on Q1 with short cct.

  • Zener power is a smaller issue and can operate at half power.

  • hFE is a also a bigger issue for max Rbias / load R ratio. so 250 is much better than 100

Most of the power is dissipated in the transistor, then R ( in short circuit) and then Zener ( no load)

Assume short cct output max current of 1A then what R value = 1/4W for a 1/2 W part.

  • (1) choose R near Zener rated voltage for 5mA or a bit more for accuracy ( not as asked for safe limit !! )
  • (2) Use Re = 0.55V / I rated (NOT 0.7V) as cutoff base current is low. Iout/hfe^2

Below 1st example is bad during a short circuit and will burn 1/4W with 2.3W

  • 2nd example does not. by using near 5mA Zener bias current.

  • using near 5mA typ datasheet rated current for Zener voltage of 13V. enter image description here

Tony Stewart EE75
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